A piece of wire m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How should the wire be cut for the square so that the total area enclosed is a minimum? Round your answer to the nearest hundredth.

Question

A piece of wire  m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How should the wire be cut for the square so that the total area enclosed is a minimum? Round your answer to the nearest hundredth.

Quizplus · Accepted Answer

Let $s$ be the length of the wire used for the square, and $t$ be the length of the wire used for the equilateral triangle. Then $s+t=11$, and the area of the square is $\left(\frac{s}{4}ight)^2=\frac{s^2}{16}$, while the area of the equilateral triangle is $\frac{\sqrt{3}}{4}t^2$. Using the fact that $s=11-t$ to eliminate $s$, we find that the total area enclosed by the figures is $$\frac{(11-t)^2}{16}+\frac{\sqrt{3}}{4}t^2.$$ Taking the derivative of this expression and setting it to 0, we find that the value of $t$ that minimizes the area is $$t=\frac{4}{3}\sqrt{3},$$and hence the length of wire used for the square is $$s=11-\frac{4}{3}\sqrt{3}\approx 6.61.$$Therefore, the wire should be cut into pieces of length approximately 6.61 and 4.39 meters for the square and equilateral triangle, respectively. 
Rounding the answer to the nearest hundredth gives 6.61, which corresponds to choice $\boxed{	extbf{(B)}}$.

A Piece of Wire M Long Is Cut into Two